The positive integers are to be partitioned into several sets such that
two times an integer and the integer itself do not belong to the same
set, and every positive integer belongs to exactly one set. What is the
minimum number of such sets required?

7 comments

You need two sets. To find which set an integer belongs in, find the
largest power of two which can divide the integer to produce a new
integer. Put the integers which have odd largest power of two divisor in
one set and the rest in the other set.

Tim Ophelders solved this puzzle:

Let \(m(n)\) be the power of \(2\) in the prime factorization of \(n\).
Define two sets \(A\) and \(B\) such that a positive integer \(x \in A\)
if and only if \(m(x)\) is even and \(x \in B\) otherwise. Then every
positive integer belongs to one of those sets. Now, for each positive
integer \(x\), their double, \(2x\), must be in the other set because
\(m(2x) = 1 + m(x)\). Thus, the minimum number of sets required to
separate every positive integer from its double is \(2\).

Simon solved this puzzle:

Write each natural number in its prime factorization form. In the first
set are the integers where the prime factorization has an even
number of \(2\)s. In the second set are the integers where the prime
factorization has an odd number of \(2\)s. Any number in the first set
will have it's double in the second set, and vice versa. Therefore the
minimum number of such sets required is (2).

It seems two sets should suffice. The following construction shows this.
Let the sets be \(A\) and \(B\). Let \(1\) be in set \(A\). For every
successive positive integer, we check if half of it is an integer. If
not, put it in set \(A\). If half the number is an integer, put it in
the set which does not have that integer. This way, every integer gets
placed in a set, and its double is never in the same set.

Alternately, every positive integer can be represented uniquely as \(2^m
(2n + 1)\), for non-negative integers \(m\) and \(n\). Let set \(A\)
contain all the positive integers where \(m\) is even, and set B contain
all integers where \(m\) is odd. It may be noted that doubling changes
the parity of \(m\), hence each integer and it's double will be in
different sets.

Two sets are clearly required so that \(1\) and \(2\) for instance lie in
different sets. Two sets are also sufficient. Here is how to do it. Call
the sets \(s_1\) and \(s_2\). For each integer \(i\), let \(2^{t_i}\) be
the largest power of \(2\) that divides it. If \(t_i\) is odd, put \(i\)
in \(s_1\), else, put it in \(s_2\).

The minimum number of sets required is two. We can split the integers
according to how many powers of \(2\) occur in each one's prime
factorization. Integers with an even power of \(2\) in their prime
factorization (including 2^0, aka odd numbers) go in the first set.
Integers with an odd power of \(2\) go in the second set. Now the double
of any integer in the first set occurs in the second set, and vice
versa. Finally, we know that we can do this with all the positive
integers because each integer either has an odd or even power of two in
its prime factorization. In addition, we know the sets are mutually
exclusive because an integer cannot both have an odd and even power of
two.

David Robertson solved this puzzle:

Every positive integer can be divided by \(2\) repeatedly until division
is no longer possible. Count the number of maximum possible divisions
(possibly zero) for each integer. Then place integers divisible by two
an odd number of times into one set, and the rest (each of which 2
divides into an even number of times) into another. By construction,
neither set contains both an element and its double.

A one-set 'partition' would fail (as it would contain \(1\) and \(2\), for
example) and so the minimum number of sets required for the partition is
\(2\).